How can I find $E_1$ and $E_1^{-1}$?

Question

Suppose that: $E_1 \begin{bmatrix}12\\35\end{bmatrix} = \begin{bmatrix}48\\35\end{bmatrix}$ Find $E_1$ and $E_1^{-1}$

I know how to find the inverse of a matrix, that's easy. It's just $E_1^{-1} = E_1 * I$

What I don't understand is how I can find $E_1$?

I know the equation in my book says $A\vec{x} = b$

Am I essentially finding $A$? If that is so, would I need to find the inverse of $x$?

So would I need to manipulate $Ax = b$ to $x^{-1}[Ax = b] = AI = x^{-1}b \implies A = x^{-1}b$?

Answer 1

Note: this answer was written for the original form of the problem $E_1\begin{bmatrix}12\\35\end{bmatrix} = \begin{bmatrix}48\\35\end{bmatrix}$. The recent edit changed the numbers slightly, however the method is unchanged, just some numbers will be different in the end. The final result is that the problem of finding $E_1$ is essentially one of a system of two equations and four unknowns, which is underdetermined. If any solution exists, it will be infinitely many solutions. Since the edit did not sufficiently change the problem, I will not update my answer to reflect the current numbers.

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Let $E_1=\begin{bmatrix}a&b\\c&d\end{bmatrix}$

The given conditions can be reworded as:

$\begin{cases} 12a+35b = 48\\ 12c+35d = 35\end{cases}$

Letting $a$ be anything, we can describe $b$ in terms of $a$. Similarly, letting $c$ be anything, we can describe $d$ in terms of $c$.

Then, $E_1 = \begin{bmatrix}a&(\frac{48-12a}{35})\\ c&(\frac{35-12c}{35})\end{bmatrix}$

In the case that the determinant of $E_1$ is nonzero, then it will have an inverse.

$E_1^{-1} = \frac{1}{a(\frac{35-12c}{35})-c(\frac{48-12a}{35})}\begin{bmatrix}(\frac{35-12c}{35})&-(\frac{48-12a}{35})\\-c&a\end{bmatrix}$

Any choice of $a$ and any choice of $c$ such that the determinant is nonzero will form a matrix $E_1$ satisfying the desired conditions. Hence, there are infinitely many solutions.

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For example, with $a=c=1$ we have: $E_1 = \begin{bmatrix}1&\frac{36}{35}\\1&\frac{13}{35}\end{bmatrix}$

For another example, with $a=2$ and $c=-1$ we have $E_1 = \begin{bmatrix}2&\frac{24}{35}\\-1&\frac{47}{35}\end{bmatrix}$

Notice that in both of those cases, we have $E_1\cdot \begin{bmatrix}12\\35\end{bmatrix}=\begin{bmatrix}48\\35\end{bmatrix}$

Answer 2

There's not enough information to uniquely determine an $E_1$. You can find one such $E_1$ that works by just setting the first column arbitrarily and solving for the second.

Also, $E^{-1}$ is the inverse of $E_1$. It is not true that $E^{-1} = (E_1)(I) = E_1$.